Elementary Principles 489 - 11.23 12.0 kg/min 25oC 12.0...

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11- 18 11.23 a. Energy Balance: MC dT dt mC T UA T T vp =− + ± 25 bg b g steam M m CC UA T dT dt T t T = = ≈= ° =⋅ ° = = 760 12 0 230 115 167 8 150 00224 0 25 kg kg min oo kJ (min C) kJ (min C) sat'd; 7.5bars C C C steam ± . . . . /.. ( m i n ) ,, af b. Steady State: dT dt TT ss == ⇒ = ° 0 150 0 0224 67 .. C c. dT T dt t T T t t T f 150 0 0224 1 0 0224 150 0 0224 094 150 094 0 0224 0 0224 0 25 . ln . e x p ( .) . =⇒ = F H G I K J ⇒= −− z z tT = ° 40 498 min. C . d. U changed. Let xU A new = () . The differential equation becomes: dT dt xx T =+− + 0 3947 0 096 0 01579 5721 ( . . ) kJ / (min C) o dT T dt x x 0 3947 0 096 0 01579 5721 10 1 0 01579 5721 10 0 3947 0 096 0 01579 5721 10 55 0 3947
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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